2x+3y-4y^2+1=0

Lösung

2 x + 3 y - 4 y^{2} + 1 = 0

(h, k) = (- \frac{25}{32}, \frac{3}{8}), p = \frac{1}{8}

Schritte zur Lösung

2 x + 3 y - 4 y^{2} + 1 = 0

Rewrite

2 x + 3 y - 4 y^{2} + 1 = 0

in the standard form

4 \cdot \frac{1}{8} (x - (- \frac{25}{32})) = (y - \frac{3}{8})^{2}

Therefore parabola properties are:

(h, k) = (- \frac{25}{32}, \frac{3}{8}), p = \frac{1}{8}

2 x + y - 10 = y^{2} - 8 x - 4

x = - 3 (y + 2)^{2} + 1

(x - 4)^{2} = 8 (y + 2)

a^{2} + b = 0

y^{2} = - 2 (x - 9)